3.71 \(\int \frac {1}{\cos ^{-1}(a x)^4} \, dx\)

Optimal. Leaf size=78 \[ -\frac {\sqrt {1-a^2 x^2}}{6 a \cos ^{-1}(a x)}+\frac {\sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^3}+\frac {\text {Ci}\left (\cos ^{-1}(a x)\right )}{6 a}+\frac {x}{6 \cos ^{-1}(a x)^2} \]

[Out]

1/6*x/arccos(a*x)^2+1/6*Ci(arccos(a*x))/a+1/3*(-a^2*x^2+1)^(1/2)/a/arccos(a*x)^3-1/6*(-a^2*x^2+1)^(1/2)/a/arcc
os(a*x)

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Rubi [A]  time = 0.16, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {4622, 4720, 4724, 3302} \[ -\frac {\sqrt {1-a^2 x^2}}{6 a \cos ^{-1}(a x)}+\frac {\sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^3}+\frac {\text {CosIntegral}\left (\cos ^{-1}(a x)\right )}{6 a}+\frac {x}{6 \cos ^{-1}(a x)^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[a*x]^(-4),x]

[Out]

Sqrt[1 - a^2*x^2]/(3*a*ArcCos[a*x]^3) + x/(6*ArcCos[a*x]^2) - Sqrt[1 - a^2*x^2]/(6*a*ArcCos[a*x]) + CosIntegra
l[ArcCos[a*x]]/(6*a)

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 4622

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(Sqrt[1 - c^2*x^2]*(a + b*ArcCos[c*x])^(n + 1)
)/(b*c*(n + 1)), x] - Dist[c/(b*(n + 1)), Int[(x*(a + b*ArcCos[c*x])^(n + 1))/Sqrt[1 - c^2*x^2], x], x] /; Fre
eQ[{a, b, c}, x] && LtQ[n, -1]

Rule 4720

Int[(((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> -Simp
[((f*x)^m*(a + b*ArcCos[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] + Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), Int[(f*x)
^(m - 1)*(a + b*ArcCos[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && LtQ[n,
-1] && GtQ[d, 0]

Rule 4724

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> -Dist[d^p/c^
(m + 1), Subst[Int[(a + b*x)^n*Cos[x]^m*Sin[x]^(2*p + 1), x], x, ArcCos[c*x]], x] /; FreeQ[{a, b, c, d, e, n},
x] && EqQ[c^2*d + e, 0] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 0])

Rubi steps

\begin {align*} \int \frac {1}{\cos ^{-1}(a x)^4} \, dx &=\frac {\sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^3}+\frac {1}{3} a \int \frac {x}{\sqrt {1-a^2 x^2} \cos ^{-1}(a x)^3} \, dx\\ &=\frac {\sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^3}+\frac {x}{6 \cos ^{-1}(a x)^2}-\frac {1}{6} \int \frac {1}{\cos ^{-1}(a x)^2} \, dx\\ &=\frac {\sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^3}+\frac {x}{6 \cos ^{-1}(a x)^2}-\frac {\sqrt {1-a^2 x^2}}{6 a \cos ^{-1}(a x)}-\frac {1}{6} a \int \frac {x}{\sqrt {1-a^2 x^2} \cos ^{-1}(a x)} \, dx\\ &=\frac {\sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^3}+\frac {x}{6 \cos ^{-1}(a x)^2}-\frac {\sqrt {1-a^2 x^2}}{6 a \cos ^{-1}(a x)}+\frac {\operatorname {Subst}\left (\int \frac {\cos (x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{6 a}\\ &=\frac {\sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^3}+\frac {x}{6 \cos ^{-1}(a x)^2}-\frac {\sqrt {1-a^2 x^2}}{6 a \cos ^{-1}(a x)}+\frac {\text {Ci}\left (\cos ^{-1}(a x)\right )}{6 a}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 71, normalized size = 0.91 \[ \frac {2 \sqrt {1-a^2 x^2}-\sqrt {1-a^2 x^2} \cos ^{-1}(a x)^2+\cos ^{-1}(a x)^3 \text {Ci}\left (\cos ^{-1}(a x)\right )+a x \cos ^{-1}(a x)}{6 a \cos ^{-1}(a x)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCos[a*x]^(-4),x]

[Out]

(2*Sqrt[1 - a^2*x^2] + a*x*ArcCos[a*x] - Sqrt[1 - a^2*x^2]*ArcCos[a*x]^2 + ArcCos[a*x]^3*CosIntegral[ArcCos[a*
x]])/(6*a*ArcCos[a*x]^3)

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{\arccos \left (a x\right )^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccos(a*x)^4,x, algorithm="fricas")

[Out]

integral(arccos(a*x)^(-4), x)

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giac [A]  time = 0.19, size = 66, normalized size = 0.85 \[ \frac {\operatorname {Ci}\left (\arccos \left (a x\right )\right )}{6 \, a} + \frac {x}{6 \, \arccos \left (a x\right )^{2}} - \frac {\sqrt {-a^{2} x^{2} + 1}}{6 \, a \arccos \left (a x\right )} + \frac {\sqrt {-a^{2} x^{2} + 1}}{3 \, a \arccos \left (a x\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccos(a*x)^4,x, algorithm="giac")

[Out]

1/6*cos_integral(arccos(a*x))/a + 1/6*x/arccos(a*x)^2 - 1/6*sqrt(-a^2*x^2 + 1)/(a*arccos(a*x)) + 1/3*sqrt(-a^2
*x^2 + 1)/(a*arccos(a*x)^3)

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maple [A]  time = 0.04, size = 63, normalized size = 0.81 \[ \frac {\frac {\sqrt {-a^{2} x^{2}+1}}{3 \arccos \left (a x \right )^{3}}+\frac {a x}{6 \arccos \left (a x \right )^{2}}-\frac {\sqrt {-a^{2} x^{2}+1}}{6 \arccos \left (a x \right )}+\frac {\Ci \left (\arccos \left (a x \right )\right )}{6}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/arccos(a*x)^4,x)

[Out]

1/a*(1/3/arccos(a*x)^3*(-a^2*x^2+1)^(1/2)+1/6*a*x/arccos(a*x)^2-1/6/arccos(a*x)*(-a^2*x^2+1)^(1/2)+1/6*Ci(arcc
os(a*x)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {a^{2} \arctan \left (\sqrt {a x + 1} \sqrt {-a x + 1}, a x\right )^{3} \int \frac {\sqrt {a x + 1} \sqrt {-a x + 1} x}{{\left (a^{2} x^{2} - 1\right )} \arctan \left (\sqrt {a x + 1} \sqrt {-a x + 1}, a x\right )}\,{d x} + a x \arctan \left (\sqrt {a x + 1} \sqrt {-a x + 1}, a x\right ) - \sqrt {a x + 1} \sqrt {-a x + 1} {\left (\arctan \left (\sqrt {a x + 1} \sqrt {-a x + 1}, a x\right )^{2} - 2\right )}}{6 \, a \arctan \left (\sqrt {a x + 1} \sqrt {-a x + 1}, a x\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccos(a*x)^4,x, algorithm="maxima")

[Out]

1/6*(6*a^2*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^3*integrate(1/6*sqrt(a*x + 1)*sqrt(-a*x + 1)*x/((a^2*x^2
- 1)*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)), x) + a*x*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x) - sqrt(
a*x + 1)*sqrt(-a*x + 1)*(arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^2 - 2))/(a*arctan2(sqrt(a*x + 1)*sqrt(-a*x
+ 1), a*x)^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\mathrm {acos}\left (a\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/acos(a*x)^4,x)

[Out]

int(1/acos(a*x)^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\operatorname {acos}^{4}{\left (a x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/acos(a*x)**4,x)

[Out]

Integral(acos(a*x)**(-4), x)

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